Question 636·Medium·Algebra

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The linear equation $(k + 2)x + 6 = 6$ is required to have infinitely many solutions. What value of $k$ satisfies this condition?

The cue is infinitely many solutions: the linear equation must be an identity, true for every $x$ . Compare the terms containing $x$ and the plain numbers on both sides. If the plain numbers already match, make the coefficient of $x$ zero, then find the missing constant. Testing only $x=0$ is a trap because it can work even when other values don't.

Hints

Hint 1
An identity is true for every value of $x$ . The plain numbers on the two sides already match. What must happen to the term containing $x$ so changing $x$ cannot change the equality?
Hint 2
The coefficient of $x$ is the number multiplying it. Make that coefficient zero, then find $k$ . If you use a Desmos regression on the original equation, choose $x=1$ : testing $x=0$ hides the coefficient.

Step-by-step

Make the two sides identical
Step 1
Match the terms containing x
Infinitely many solutions means the equation must work for every $x$ . The plain numbers are already $6$ on both sides, so the coefficient, the number multiplying $x$ , must be $0$ :
$k+2=0.$
Step 2
Check why zero gives infinitely many solutions
If $k+2=0$ , then $0x=0$ for every $x$ . The original equation becomes
$6=6.$
That's true for every $x$ , not a no-solution case. When the $x$ -terms disappear, check whether the plain numbers still match.
Step 3
Find the value of k
Type $(k_1+2)(1)+6\sim6$ in Desmos. Because every $x$ must work, $x=1$ is a valid test; $x=0$ would hide the term containing $k$ . The $\sim$ asks Desmos to fit $k_1$ rather than make a $k$ slider. Under PARAMETERS, it shows $k_1=-2$ , so $k=-2$ makes the equation true for every $x$ . Choice D.

Question 636·Medium·Algebra

Strategy

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Step-by-step

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